Question

Question: An aluminum kettle weighs 1.05 kg.

(a) What is the heat capacity of the kettle?

(b) How much heat is required to increase the temperature of this kettle from 23.0 °C to 99.0 °C?

(c) How much heat is required to heat this kettle from 23.0 °C to 99.0 °C if it contains 1.25 L of water (density of 0.997 g/mL and specific heat of 4.184 J/g °C)?

Short answer

1. The heat capacity of the kettle is941.85\(\frac{J}{{^0C}}\).
2. The heat required to increase the temperature of the kettle = 71.58 kJ.

3. The heat required to raise the temperature of the kettle with water inside it is 4048kJ.

Step-by-step solution

Step 1: Heat capacity

The heat capacity of a substance can be calculated by the formula H = mc, where “m” is the mass of the substance and “c” is its specific heat capacity.

Step 2:Heat capacity of an aluminum kettle in joules

(a) We know that the specific heat capacity of aluminum is0.897J/g℃.

The mass of the aluminum kettle = 1.05 kg = 1,050 g (given).

Therefore, the heat capacity of the aluminum kettle in joules = 1050\( \times \) 0.897 = 941.85\(\frac{J}{{^0C}}\).

Specific heat

The heat required to raise the temperature of a substance is given by the formulaQ = C × m ×∆ T,where “C” is the specific heat of the substance, “m” is the mass of the substance, and “∆T” is the change in the temperature of the substance.

Amount of heat required

(b) We know from the given details that:

C = 0.897 J/g °C(From table 5.1)

m = 1050 g

T = change in temperature = \({T_{final}} - {T_{initial}} = {99^0}C - {23^0}C = {76^0}C\)

By putting the values above in the equation Q = C × m × ∆ T, we get:

\(\begin{array}{c}\;Q{\rm{ }} = {\rm{ }}0.897 \times 1050 \times 76{\rm{ }}\\ = {\rm{ }}71580{\rm{ }}J{\rm{ }}\\ = {\rm{ }}71.58{\rm{ }}kJ\end{array}\)

The heat required to raise the temperature of the aluminum kettle is 71.58 kJ.

Amount of heat required if the kettle contains 1.25 L of water

(c) We know from the given details that:

C\(_{alu\min um}\) = 0.897 J/g °C(From table 5.1)

m\(_{alu\min um}\) = 1050 g

C\(_{water}\)= 4.186 J/g °C

The density of the water = 1g/mL

The volume of the water = 1.25 L = 1250 mL

The mass of the water = density of water \( \times \)volume of water

m\(_{water}\)= 12500 g .

T = change in temperature = 76\(^0C\).

The known values are substituted as shown below:

Q = \({m_{alu\min um}} \times {C_{alu\min um}} \times \Delta T + {m_{water}} \times {C_{water}} \times \Delta T\)

Q = 1050 \( \times \) 0.897 \( \times \)76 + 12500 \( \times \) 4.186 \( \times \) 76

Q = (71580 + 3976700) J = 4048280 J = 4048 kJ

Thus, the heat required to raise the temperature of the aluminum kettle with water in it is 4048 kJ.